Answer:
Acceleration, 
Explanation:
Initial speed of the skater, u = 8.4 m/s
Final speed of the skater, v = 6.5 m/s
It hits a 5.7 m wide patch of rough ice, s = 5.7 m
We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :
So, the acceleration on the rough ice 
and negative sign shows deceleration.
Answer:
there are 7 significant figures
Explanation:
15.33879+15.555
=30.89379
there are 7 significant figures
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Answer:
Explanation:
Horizontal displacement
x = 120 t
Vertical position
y = 3610 - 4.9 t²
y = 0 for the ground
0 = 3610 - 4.9 t²
t = 27.14 s
This is the time it will take to reach the ground .
During this period , horizontal displacement
x = 120 x 27.14 m
= 3256.8 m
So packet should be released 3256.8 m before the target.
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